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btw, the question is: \int_{0}^{4} x\sqrt{16-x^2} dx And then at the end it's like: "where x=4sin(\theta)" for some reason

Uhh, why can't you use the substitution u = 16-x^2

Probably not.

Depends on how well your school goes, and how well you go...

Dear current #3 secret crush, Thank you for finding me those people to have a chat with - it was really appreciated =)

I do talk to girls in my lectures...? That's how I have my English lecture/tute buddies...?

what i go to arts lectures because i'm an actual arts student. but uhh you know... that's nice. you should try talking to them.

How so?

Re: MX2 Integration Marathon In what? Complex analysis. It's fascinating. Obviously not going to write on... second year topics. But if the deeper stuff is just as fun and great as what I'm learning now - then... why not? It should be awesome fun

Yeah, pretty much. Arts students. gotta love 'em.

Arts students. aha knew it

what lecture was this (no crepe)

huh what

Re: MX2 Integration Marathon Well, my method uses integrating over a contour in the complex plane and taking real and imaginary parts. Which is pretty cool because... well, it is. if i survive until honours - i might like to dabble in this sort of thing.

The Grill. Quad Food Court.

\int_{0}^{\infty} cos(x^{2}) dx = \frac{\sqrt{\pi}}{2\sqrt{2}}

Re: MX2 Integration Marathon It's pretty cool how it's derived. Difficult, but cool. More cool once you look at the graph: http://www.wolframalpha.com/input/?i=cos%28x^2%29+from+0+to+infinity It's the area from 0 to whatever the first root is, then minus the area, then plus a little, minus...

Re: The Nonsense Thread From today? No, there are reasons why I wasn't in the best mood today... but no, I have no problem with mirakon.

Re: MX2 Integration Marathon oh oops, limits are 0 to infinity. LOLOLOLOL my bad \int_{0}^{\infty} cos(x^{2}) dx = \frac{\sqrt{\pi}}{2\sqrt{2}}

Re: The Nonsense Thread taller than expected